Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 10 - Interactions and Potential Energy - Exercises and Problems: 9

Answer

(a) The ball's speed at the lowest point is 1.4 m/s (b) The pendulum will swing up to an angle of $30^{\circ}$ on the other side.

Work Step by Step

(a) Let $L$ be the length of the pendulum. We can find the height $h$ (above the ball's lowest point) when the pendulum is at a $30^{\circ}$ angle. $\frac{L-h}{L} = cos(\theta)$ $h = L~[1-cos(\theta)]$ $h = (0.75~m)~[1-cos(30^{\circ})]$ $h = 0.10~m$ We can use conservation of energy to find the ball's speed at the lowest point. The kinetic energy at the lowest point will be equal to the potential energy at height $h$. So; $KE = PE$ $\frac{1}{2}mv^2 = mgh$ $v^2 = 2gh$ $v = \sqrt{2gh}$ $v = \sqrt{(2)(9.80~m/s^2)(0.10~m)}$ $v = 1.4~m/s$ The ball's speed at the lowest point is 1.4 m/s (b) By conservation of energy, the ball will go up to the same height $h = 0.10~m$ on the other side. Therefore, the pendulum will swing up to an angle of $30^{\circ}$ on the other side.
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